Question:
The solution of the differential equation \(\displaystyle \frac{dy}{dx}=1+y^2\) is
The solution of the differential equation \(\displaystyle \frac{dy}{dx}=1+y^2\) is
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For \(\frac{dy}{dx}=1+y^2\), separate variables and use \(\int\frac{dy}{1+y^2}=\tan^{-1}y\).
Updated On: May 7, 2026
- \(y=\tan x+c\)
- \(y=\tan(x+c)\)
- \(y=\tan x\)
- \(y=-\tan(x+c)\)
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The Correct Option is B
Solution and Explanation
We are given:
\[
\frac{dy}{dx}=1+y^2.
\]
This equation can be separated as:
\[
\frac{dy}{1+y^2}=dx.
\]
Now integrate both sides:
\[
\int \frac{dy}{1+y^2}=\int dx.
\]
We know:
\[
\int \frac{dy}{1+y^2}=\tan^{-1}y.
\]
Also,
\[
\int dx=x+c.
\]
Therefore,
\[
\tan^{-1}y=x+c.
\]
Now take tangent on both sides:
\[
y=\tan(x+c).
\]
Hence, the solution is:
\[
y=\tan(x+c).
\]
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